题目内容：

AD是△ABC中BC边上的高,在AD上取点E,使AE=1/2ED,过E作直线MN//BC,交AB于M,交AC于N,现将△AMN沿MN折过去,此时A点到了A1点的位置,成了立体图形,如果设角A1ED=60度,求证 EA1⊥平面A1BC

优质解答

证明：连结A1D∵AD⊥BC MN//BC △AMN沿MN折∴BC⊥EA1 BC⊥ED EA1∩ED于E∴BC⊥面A1ED而EA1∈面A1ED∴BC⊥EA1∵设EA1=1,则ED=2EA1=2而∠A1ED=60° ∴A1D^2=EA1^2+ED^2-2EA1·ED·cos∠A1ED=1^2+2^2-2×1×2×cos∠60...